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KengaRu [80]
3 years ago
13

Summarize the procedures you perform before, during, and after a lab investigation.

Physics
2 answers:
natka813 [3]3 years ago
5 0
What the person Before me said



.........
Gelneren [198K]3 years ago
4 0

Answer:

Before Lab

At all times while in the lab area you must follow all classroom rules and behave in an appropriate manner or

you will not be allowed to participate in laboratory activities. You should come to class prepared with any

pre-lab assignments or instructions that have been given.

⋅ While working in the lab area safety goggles should be worn at all times.

⋅ All long hair should be tied back; open-toed shoes, baggy clothing, and loose jewelry should also be

avoided.

⋅ Know the location and proper use of all safety equipment (including fire extinguisher, fire blanket, firstaid kit, eye wash station, and sink) as well as classroom exits.

⋅ Food and drink are not allowed in the lab area.

⋅ Never work alone in the laboratory.

⋅ If there is an accident or a problem, immediately stop what you are doing and report it to your instructor.

During Lab

⋅ Before beginning any procedures, read through all lab instructions that your instructor has given.

⋅ If you do not understand the instructions or the procedures, ask for clarification.

⋅ Perform only procedures or experiments that have been assigned by your instructor.

⋅ Clean up all spills, including water, and messes immediately before proceeding.

⋅ Wear an apron whenever using acids, bases, or stains.

⋅ All solutions, chemicals, and specimens should be treated with care.

⋅ Never taste or smell anything in the lab. If instructed to smell something, use your hand to gently waft the vapors to

your nose.

Fire / Electrical Safety

⋅ Know the location and proper use of the fire blanket and fire extinguisher.

⋅ Never directly touch something that may be hot, use a hot mitt or tongs instead.

⋅ When heating, always point test tubes away from yourself or others.

⋅ Never use electrical equipment near water or moisture.

⋅ Extinguish all uncontrolled flames immediately and notify your instructor.

⋅ If a burn is received, immediately place the burned area under cold water and notify instructor.

⋅ Always make sure that all flames and electrical equipment are turned off when not in use.

Glassware / Sharp Object Safety

⋅ Never use glassware that is cracked or chipped (even if it does not leak).

⋅ When using a blade, never use a dull blade and always cut downward and away from your body.

⋅ If a cut is received, immediately wash the cut, apply pressure to stop the bleeding, and notify your

instructor.

⋅ Always dispose of broken glass in a glass disposal box and notify your instructor.

Chemical Safety

⋅ Know the location and proper use of the eye-wash, sinks, and chemical showers.

⋅ Know the location of sand or vermiculite in case of acid spills.

⋅ When diluting, always pour acids into water. Do not pour water into acid.

⋅ If a chemical is spilled on your person, immediately flush the area with water and notify the instructor.

⋅ Dispose of chemicals only as instructed in marked containers.

Biological Safety

⋅ Wear protective gloves when handling biological specimens.

⋅ If you come into direct contact with biological specimens or fluids, immediately notify your instructor.

⋅ Dispose of all biological cultures or materials containing live cultures in the proper biohazard bags.

After Lab

Before leaving the lab area, return all equipment to its proper place and make sure that your lab area is

cleaned up. After completing the lab wash your hands thoroughly with soap and water.

(Hope this helped)

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Answer:

v_m \approx -4.38\; \rm m \cdot s^{-1} (moving toward the incline.)

v_M \approx 4.02\; \rm m \cdot s^{-1} (moving away from the incline.)

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Explanation:

If g = 9.81\; \rm m \cdot s^{-2}, the potential energy of the block of m = 2.20\; \rm kg would be m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this m = 2.20\; \rm kg\! block right before the collision would also be approximately 77.695\; \rm J.

Calculate the velocity of that m = 2.20\; \rm kg based on its kinetic energy:

\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}.

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right before the collision: approximately 18.489\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right after the collision: (m\cdot v_m + m \cdot v_M).

For momentum to conserve in this collision, v_m and v_M should ensure that m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

Kinetic energy of the two blocks right before the collision: approximately 77.695\; \rm J and 0\; \rm J. Sum of these two values: approximately 77.695\; \rm J\!.

Sum of the energy of each block right after the collision:

\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right).

Similarly, for kinetic energy to conserve in this collision, v_m and v_M should ensure that \displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J.

Combine to obtain two equations about v_m and v_M (given that m = 2.20\; \rm kg whereas M = 7.00\; \rm kg.)

\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right..

Solve for v_m and v_M (ignore the root where v_M = 0.)

\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right..

The collision flipped the sign of the velocity of the m = 2.20\; \rm kg block. In other words, this block is moving backwards towards the incline after the collision.

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